Question

若函数f(x)=1/3x^3-a^2x满足对于任意的x1,x2属于[0,1]都有|f(x1)-f(x2)|<=1恒成立，求a的取值范围

Answer 1

f(x1)-f(x2)
=(1/3) [(x1-x2)(x1^2 +x1·x2 +x2^2)]-(a^2)(x1-x2)
=(1/3)(x1-x2)(x1^2 +x1·x2 +x2^2 -3a^2)
|f(x1)-f(x2)|≤1
则|(1/3)(x1-x2)(x1^2 +x1·x2 +x2^2 -3a^2)|≤1;
|x1-x2|·|x1^2 +x1·x2 +x2^2 -3a^2|≤3;
而|x1-x2|≤1,则如果满足|x1^2 +x1·x2 +x2^2 -3a^2|≤3,则
|x1-x2|·|x1^2 +x1·x2 +x2^2 -3a^2|≤3即|f(x1)-f(x2)|≤1恒成立。

x1^2 +x1·x2 +x2^2 -3a^2
=(x1-x2)^2+3x1·x2 -3a^2
由于x1,x2属于[0,1]，则 3-3a^2≥(x1-x2)^2+3x1·x2 -3a^2≥ -3a^2;
则由|x1^2 +x1·x2 +x2^2 -3a^2|≤3得: -3a^2≥-3;且3-3a^2≤3
解得: -1≤a≤1.

Answer 2

现将函数求导，不等式转化成最大值减最小值小于等于1，然后带着参数分类讨论